Question

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Answer 1

EXPLANATION.

Bisector of ∠B and ∠C of a ΔABC.

Intersect at at O.

To prove :

⇒ ∠BOC = 90° + ∠A/2.

As we know that,

Sum of the angles of a triangle is equal to 180°.

In ΔABC.

⇒ ∠A + ∠B + ∠C = 180°.

⇒ ∠A/2 + ∠B/2 + ∠C/2 = 180°/2.

⇒ ∠A/2 + ∠1 + ∠2 = 90°.

⇒ ∠1 + ∠2 = 90° - ∠A/2. - - - - - (1).

In ΔOBC.

⇒ ∠BOC + ∠OBC + ∠OCB = 180°.

⇒ ∠BOC + ∠1 + ∠2 = 180°.

⇒ ∠BOC + 90° - ∠A/2 = 180°.

⇒ ∠BOC = 180° - 90° + ∠A/2.

⇒ ∠BOC = 90° + ∠A/2.

Hence Proved.

Answer 2

Answer:

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